 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ia/paper-4/6e/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Suppose

x^{2} \equiv - 1 (mod p)

. Then

p ∤ x

x^{p - 1} \equiv 1

. But

p \equiv 3 (mod 4)

makes

(p - 1) /2

odd, and hence

x^{p - 1} = (x^{2})^{(p - 1) /2} \equiv (- 1)^{(p - 1) /2} = - 1 (mod p) .

Since this

p

is odd,

1 \neq \equiv - 1 (mod p)

, a contradiction. There is no square root of minus one in this case.

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